Shaded area
= 3 x 3 x 4 + 6 x 6 + 6 x 2 + 6 x 4
= 108cm²

There are 4 small parallelograms of base 3cm x perpendicular height 3cm.

Area of parallelogram = base x perpendicular height
(Or area of two triangles inside it divided by a diagonal = ½ x base x height x 2 )

The left and right shaded areas 2 congruent trapeziums. When combined, they give a square of 6cm by 6cm.

The top and bottom shaded areas are also parallelograms with base of 6cm. One has perpendicular height 4cm the other has perpendicular height 6cm.

PQR is within a rectangle 4 squares wide and 6 squares long.

Rectangle area = 4 x 6 x 2cm x 2cm
= 96cm²

Subtract the areas of the 3 triangles bounding it.

½ x 6 x 3 + ½ x 4 x 1 + ½ x 5 x 1
= 9 + 2 + 2.5
= 13.5

13.5 x 2cm x 2cm = 54cm²

Triangle PQR area = 96cm² - 54cm² = 42cm²

Total visible area is ½ shaded ½ white.

US'V = white area PQRVS'U

Since US'V = USV, then all 3 areas are equal.

So area of each = 12 ÷ 3 = 4

Square US'VS = 4 x 2 = 8

Area of square = ½ x diagonal² = 8
Diagonal² = 16
Diagonal UV = √16 = 4

Thank you so much

https://mindyourdecisions.com/blog/2016/09/04/the-hardest-easy-geometry-problem-sunday-puzzle/

For the triangle PQR question


Edit :

PST = 35°

Draw the triangle and you will find that there are 2 isosceles triangles RQS and RQT.

From this, compare their sides.
RQ = RS and RQ = RT

This implies RS = RT . So RST is isosceles also. From here it is easy to find.

Working for last question :

draw lines from A and B to the intersection the two curved paths. Call the intersection C.

AC = BC = AB
(all are radius of the quarter circles, which are identical and are equal to one side of a square, 6cm)

So triangle ABC is equilateral
We have divided the unshaded area into
1 equilateral triangle and 2 sectors.

The two sectors are identical and have angle of 30° (90°of the square corners - each corner of the equilateral triangle).

They can be combined to form 1 sector of angle 60°.

Area of shaded parts

= Area of square - area of triangle - area of combined sector

= 6 x 6 - ½ab sin c - 60°/360° x πr²

= 36 - ½(6)(6) sin60° - 1/6 π(6²)

= 36 - 18(√3 / 2) - 6π

= 36 - 6π - 9√3

a = 36, b = 6, c = 9

a + b + c
= 36 + 6 + 9
= 51

https://artofproblemsolving.com/wiki/index.php/2009_AMC_12A_Problems/Problem_10

the solution for first question

Thankyou so much. PSWe are not allowed to use the sine formula for math yet only for physics. :( but I can still use 60/360 x (pi)r^2 and pythagoras theorem:)

Thx

No problem, most welcome.

Yes Pythagoras will also work to find the area of the triangle in the last question. Just that it's more steps.


I.e divide the △ into two right angled △, where each has a base of 3cm.

6² = 3² + h², where h is the height of each right angled △ (and also the height of the equilateral △)

Then h = √(6² - 3²) = √27 = √(9x3) = 3√3

Area of △= ½ x 6 x 3√3 = 9√3

But it's good that you understand how to do them now.

The PQR question is actually a primary school style math question lol

Yes it is,hehe. The two questions that puzzled me was actually the first Q and the last one. I asked for the others too because our main focus for Math Olympiad is to explain Qs in SMO round 2....as much as we can.

...when math questions become english questions with numbers plunged into it.. :))))

Ah so these are Olympiad style questions...

First one got me thinking a while as well. Was trying to see which shapes could be drawn . Tried trapezium but didn't work. But the thing is that the first question assumes that the quadrilateral is convex.

If it were a concave quadrilateral then it's different story already.

So my #Math#Is#Fun# is being a truth in a lie .....fornow....

I wonder what happens if it was a concave quadrilateral we wont get near the ans:(

What i was doing was assuming an angle was 90 degrees and checking if that would get me an integer using pythagoras theorem but i failed in doing so..
I used pythagoras theorem bcoz 5 is a Pythagorean triple and hoping I would somehow find its fit 12,13

Anyways thanks for your help and time :)

Yup wouldn't work, as 17² - 9² = 264, and 264 isn't an integer.

Most welcome, if you got any more question you can notify me

Edit : works for concave quadrilateral also.